# find the mean number of defective items in a sample of two items drawn one-by-one without replacement from an urn containing 6 items, which include 2 defective items. assume that the items are identical in shape and size.

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## From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random.

From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random. Let the random ... X (b) Mean of X (c) Variance of X

## From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random.

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## From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find Variance of X.

Click here👆to get an answer to your question ✍️ From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find Variance of X.

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## From a lot of 6 items containing 2 defective items, a sample of 4 items are drawn at random. Let the random variable X denote the number of defective items in the sample. If the sample is drawn without replacement, find Variance of X.

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Updated on : 2022-09-05

Solution Verified by Toppr

We have a lot of 6 items, 2 are defective →6−2=4 are non-defective.

P (drawing a defective item) =

6 2 = 3 1

P (drawing a non-defective item) = 1 - P (drawing a defective item) =1−

3 1 = 3 2

Note:4 items can be drawn from a lot of 6 items in

6 C 4 ways.

Let X be the random variable of drawing 2 defective items from the lot.

P(X=0)= P (no defective item in the sample) =

6 C 4 4 C 4 = 15 1

P(X=1)= P (one defective item from the lot) = P(one defective, 3 non-defective items) =(

6 C 4 2 C 1 4 C 3 )= 15 8

P(X=2)= P (two defective items from the lot) = P(2 defective, 2 non-defective bulbs) =

6 C 4 2 C 2 4 C 2 = 15 6

Therefore the required probability distribution is as follows:

X: 0 1 2

P(X): 15 1 15 8 15 6

So Variance of X will be :E(X

2 )−(E(X)) 2 E(X)=∑ i=0 2 x i P(X=x i ) →0× 15 1 +1× 15 8 +2× 15 6 = 15 20 E(X 2 )=∑ i=0 2 x i 2 P(X=x i ) →0 2 × 15 1 +1 2 × 15 8 +2 2 × 15 6 = 15 32 Variance of X= 15 32 −( 15 20 ) 2 = 45 16 Video Explanation

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## [Sample Paper] Find the mean number of defective items in a sample of

Question 27 (Choice 2) Find the mean number of defective items in a sample of two items drawn one-by-one without replacement from an urn containing 6 items, which include 2 defective items. Assume that the items are identical in shape and size. Let X : Number of defective items drawn Since ther

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## Question 27 (Choice 2) - CBSE Class 12 Sample Paper for 2023 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Oct. 19, 2022 by Teachoo

## Find the mean number of defective items in a sample of two items drawn one-by-one without replacement from an urn containing 6 items, which include 2 defective items. Assume that the items are identical in shape and size.

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Question 27 (Choice 2) Find the mean number of defective items in a sample of two items drawn one-by-one without replacement from an urn containing 6 items, which include 2 defective items. Assume that the items are identical in shape and size. Let X : Number of defective items drawn Since there are maximum of 2 defective items We can get 0, 1 or 2 defective items So, X = 0 or X = 1 or X = 2 Finding probabilities separately For X = 0 Two items drawn, without replacement, 0 are defective P(X = 0) = P(not defective) × P(not defective) = 4/6 × 3/5 = 𝟐/𝟓 For X = 1 Two items drawn, without replacement, 1 is defective P(X = 1) = P(not defective) × P(defective) + P(defective) × P(not defective) = 𝟒/𝟔 × 𝟐/𝟓 + 𝟐/𝟔 × 𝟒/𝟓 = 8/30+8/30 = 16/30 = 𝟖/𝟏𝟓 For X = 2 Two items drawn, without replacement, 2 are defective P(X = 2) = P(defective) × P(defective) = 2/6 × 1/5 = 𝟏/𝟏𝟓 So, probability distribution is Now, We need to find Mean or Expectation of X Mean or Expectation of X = E(X) = ∑_(𝒊 = 𝟏)^𝒏▒𝒙𝒊𝒑𝒊 = 0 ×2/5+1 × 8/15+2 × 1/15 = 8/15+ 2/15 = 10/15 = 𝟐/𝟑

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### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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